| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Standard +0.3 This is a straightforward quotient rule application with standard trigonometric functions. Part (a) requires careful algebraic manipulation to reach the given form, and part (b) is routine substitution and tangent line calculation. Slightly easier than average due to being a 'show that' question with the answer provided, requiring only verification rather than discovery. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Apply quotient rule with \(u = 3 + \sin 2x\), \(v = 2 + \cos 2x\), \(\frac{du}{dx} = 2\cos 2x\), \(\frac{dv}{dx} = -2\sin 2x\) | ||
| \(\frac{dy}{dx} = \frac{2\cos 2x(2 + \cos 2x) - (-2\sin 2x)(3 + \sin 2x)}{(2 + \cos 2x)^2}\) | M1 | Applying \(\frac{vu' - uv'}{v^2}\) |
| Any one term correct on numerator | A1 | |
| Fully correct (unsimplified) | A1 | |
| \(= \frac{4\cos 2x + 2\cos^2 2x + 6\sin 2x + 2\sin^2 2x}{(2 + \cos 2x)^2}\) | ||
| \(= \frac{4\cos 2x + 6\sin 2x + 2(\cos^2 2x + \sin^2 2x)}{(2 + \cos 2x)^2}\) | ||
| \(= \frac{4\cos 2x + 6\sin 2x + 2}{(2 + \cos 2x)^2}\) | A1* | For correct proof using \(\cos^2 2x + \sin^2 2x = 1\). No errors seen in working. |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| When \(x = \frac{\pi}{2}\): \(y = \frac{3 + \sin\pi}{2 + \cos\pi} = \frac{3}{1} = 3\) | B1 | \(y = 3\) |
| At \(\left(\frac{\pi}{2}, 3\right)\): \(m(\mathbf{T}) = \frac{6\sin\pi + 4\cos\pi + 2}{(2 + \cos\pi)^2} = \frac{-4+2}{1^2} = -2\) | B1 | \(m(\mathbf{T}) = -2\) |
| Either \(y - 3 = -2\left(x - \frac{\pi}{2}\right)\) or \(y = -2x + c\) with \(3 = -2\left(\frac{\pi}{2}\right) + c \Rightarrow c = 3 + \pi\) | M1 | \(y - y_1 = m\left(x - \frac{\pi}{2}\right)\) with their tangent gradient and their \(y_1\) |
| \(\mathbf{T}: y = -2x + (\pi + 3)\) | A1 | \(y = -2x + \pi + 3\) |
## Question 7:
### Part (a)
$y = \frac{3 + \sin 2x}{2 + \cos 2x}$
| Working | Mark | Guidance |
|---------|------|----------|
| Apply quotient rule with $u = 3 + \sin 2x$, $v = 2 + \cos 2x$, $\frac{du}{dx} = 2\cos 2x$, $\frac{dv}{dx} = -2\sin 2x$ | | |
| $\frac{dy}{dx} = \frac{2\cos 2x(2 + \cos 2x) - (-2\sin 2x)(3 + \sin 2x)}{(2 + \cos 2x)^2}$ | M1 | Applying $\frac{vu' - uv'}{v^2}$ |
| Any one term correct on numerator | A1 | |
| Fully correct (unsimplified) | A1 | |
| $= \frac{4\cos 2x + 2\cos^2 2x + 6\sin 2x + 2\sin^2 2x}{(2 + \cos 2x)^2}$ | | |
| $= \frac{4\cos 2x + 6\sin 2x + 2(\cos^2 2x + \sin^2 2x)}{(2 + \cos 2x)^2}$ | | |
| $= \frac{4\cos 2x + 6\sin 2x + 2}{(2 + \cos 2x)^2}$ | A1* | For correct proof using $\cos^2 2x + \sin^2 2x = 1$. No errors seen in working. |
**Total: 4 marks**
---
### Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| When $x = \frac{\pi}{2}$: $y = \frac{3 + \sin\pi}{2 + \cos\pi} = \frac{3}{1} = 3$ | B1 | $y = 3$ |
| At $\left(\frac{\pi}{2}, 3\right)$: $m(\mathbf{T}) = \frac{6\sin\pi + 4\cos\pi + 2}{(2 + \cos\pi)^2} = \frac{-4+2}{1^2} = -2$ | B1 | $m(\mathbf{T}) = -2$ |
| Either $y - 3 = -2\left(x - \frac{\pi}{2}\right)$ or $y = -2x + c$ with $3 = -2\left(\frac{\pi}{2}\right) + c \Rightarrow c = 3 + \pi$ | M1 | $y - y_1 = m\left(x - \frac{\pi}{2}\right)$ with their tangent gradient and their $y_1$ |
| $\mathbf{T}: y = -2x + (\pi + 3)$ | A1 | $y = -2x + \pi + 3$ |
**Total: 4 marks**
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { 3 + \sin 2 x } { 2 + \cos 2 x }$$
(a) Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 \sin 2 x + 4 \cos 2 x + 2 } { ( 2 + \cos 2 x ) ^ { 2 } }$$
(b) Find an equation of the tangent to $C$ at the point on $C$ where $x = \frac { \pi } { 2 }$. Write your answer in the form $y = a x + b$, where $a$ and $b$ are exact constants.\\
\hfill \mbox{\textit{Edexcel C3 2011 Q7 [8]}}